[I think this is morally the same as Nilotpal Sinha's answer, but it only uses asymptotic estimates and no integrals, so I thought I'd post it here in case someone found it useful or interesting.]

Start with the version of the Prime Number Theorem which says that $\pi(x) \sim x / \ln(x)$. Let $p_n$ denote the $n$-th prime, so in particular $\pi(p_n) = n$, and substitute in $p_n$ in the asymptotic estimate for $\pi(x)$ to get $n = \pi(p_n) \sim p_n / \ln(p_n)$, or equivalently

$p_n \sim n \ln p_n$

(If you're being nitpicky this works because the sequence $p_n$ is monotonic and unbounded.) Now take logs of both sides (again, if you're being nitpicky, this works because both $n$ and $p_n / \ln(p_n)$ are unbounded) and multiply by $n$ to get:

$n \ln p_n \sim n \ln(n) + n\ln\ln(p_n)$.

At this point, if you want just the leading term, you can show easily that $\lim_{n \to \infty} n \ln\ln(p_n) / n\ln(n) = 0$, so that in fact $p_n \sim n \ln p_n \sim n \ln n$, or you can probably keep substituting in to get finer and finer error terms.